# Differential/derivative of left translation for (matrix) Lie groups

I am very new to Lie groups and manifolds. In my self study, many times I have come across "differential (or derivative) of the left translation" (for example, here). I don't fully understand what is meant by differential/derivative here. I would appreciate any attempt to explain this concept both formally and intuitively.

For example, what does the differential/derivative here mean and how is it defined? Does the definition require charts in general (I have seen how smooth maps are defined between two manifolds)? Would it have a more convenient form in the special case of matrix Lie groups? What does it do?

Part of the answer given to this question is definitely relevant. But I am not sure where does $$dL_{g}(v) = \frac{d}{dt}\bigg|_{t=0} L_{g}\exp(tv)$$ come from? (I know that $$\exp(tv)$$ is a curve on $$G$$ that passes through the identity element at $$t=0$$ and $$v$$ is its tangent vector at the identity). Is this the definition of differential of $$L_{g}$$? Is there a backstory?

To make this question more concrete, let me set up a standard notation. Let $$G$$ be a matrix Lie group. $$L_g : G \to G$$ for any $$g \in G$$ maps an element of $$G \ni p$$ to $$L_g(p) := g p$$. Then, the differential of $$L_g$$, denoted by $$dL_g$$ is a mapping from a tangent space at any $$p \in G$$ to the tangent space at $$L_g(p)$$.

• $dL_g$ is the pushforward of the map $L_g$. How is that defined? – Oliver Jones Sep 30 '19 at 7:01
• Just to be clear, you are familiar with treating $dL_g$ as a map between tangent spaces, but you want clarification on how it can be written in terms of the exponential map? The formula you give only gives an expression for $dL_g$ as a map from $T_eG$ to $T_gG$. – Kajelad Sep 30 '19 at 19:23
• Technically, tangent vectors are derivations. In this sense, the above definition is incomplete. – Oliver Jones Sep 30 '19 at 20:11
• Thanks for the help folks. @Kajelad I wasn't familiar with pushforward - it makes sense. To make this complete, do I need to mention a point on $G$ (say, $p$) at which $d L_g$ is being calculated? In that first I need the curve that goes through $p$ at $t = 0$ and has a tangent vector $v$. In this case, $p \exp(t p^{-1} v)$ would be such a curve. Mapping this curve via $L_g$ gives $g p \exp(t p^{-1} v)$. The tangent vector at $t = 0$ is then going to be $g v$. And this holds for any $p$. Is this correct? – newbie777 Oct 2 '19 at 4:27
• @OliverJones Thanks for the help! I haven't fully studied the alternative definition of tangent vectors as derivatives. Are these two definitions equivalent and equally "formal"? Perhaps related to this question: what is allowing me here to take the "usual" derivative $\frac{d}{d t}$ from a curve such as $t \mapsto \exp(t v) \in G$ without involving charts? I guess this has something to do with the fact that $G$ is assumed to be a matrix Lie group - can you please elaborate? Thanks! – newbie777 Oct 2 '19 at 4:36

As you note, $$dL_g:T_hG\to T_{gh}G$$ (or the differential of any smooth map) is already a well-defined linear a map between tangent spaces without further reference to the group structure. In terms of velocities of curves, it is defined by: $$dL_g\left(\left.\frac{d}{dt}\gamma(t)\right|_{t=t_0}\right)=\left.\frac{d}{dt}\left(L_g(\gamma(t))\right)\right|_{t=t_0}$$ Equivalently, in terms of derivations: $$[dL_g(V_h)](f)=V_h(f\circ L_g)$$ These two definitions of tangent vectors are equivalent: we may equate every velocity with a derivation given by $$\left(\left.\frac{d}{dt}\gamma(t)\right|_{t=t_0}\right)(f)=\left.\frac{d}{dt}\left(f(\gamma(t))\right)\right|_{t=t_0}$$ If this isn't already familiar, it might be worth checking that the above definitions of the differential agree.

Understanding its relationship with the exponential map will require establishing a few facts about the Lie algebras and the exponential map.

The Lie algebra $$\text{Lie}(G)$$, can be identified with two vector spaces:

• The tangent space $$T_eG$$
• The space of left-invarant vector fields on $$G$$

To go from the first to the second, we can just evaluate the vector field at $$e$$. To go from a tangent vector $$v\in T_e M$$ to a vector field $$V$$, we let $$V_e=v$$, and the value of $$V$$ at any other point $$g$$ must be $$V_g=dL_gv$$ by left invariance. Because of this, we often use the same symbol to refer to both a tangent vectors at $$e$$ and the corresponding left-invariant vector field. Instead, I'll treat elements $$V\in\text{Lie}(G)$$ as left-invariant vector fields and use subscripts $$V_e\in T_eM$$ to refer to their values at particular points.

The exponential map $$\exp:\text{Lie}(G)\to G$$ is given by $$\exp(V)=\gamma_V(1)$$ Where $$\gamma_V$$ is the unique integral curve of the left-invariant vector field $$V$$ satisfying $$\gamma_V(0)=e$$ and $$\frac{d}{dt}\gamma(t)=V_{\gamma(t)}$$ (often called the one parameter subgroup generated by $$V$$). A consequence of this is that $$\gamma_V(t)=\exp(tV)$$.

Let $$U_e\in T_e G$$, and $$U$$ be the corresponding left-invariant vector field. We can use definition of the differential, and the fact that $$\left.\frac{d}{dt}\exp(tU)\right|_{t=0}=U_e$$ to arrive at the formula you give: $$\left.\frac{d}{dt}\left[L_g\exp(tU)\right]\right|_{t=0}=dL_g\left.\frac{d}{dt}\exp(tU)\right|_{t=0}=dL_gU_e$$ We see, somewhat more explicitly, this is only an expression for how $$dL_g$$ acts on tangent vectors at the identity.

• Thanks @Kajelad - and can you please help me to understand why are we allowed to talk about $\frac{d}{dt}$ of a curve that lives on the matrix Lie group? (as far as I know, $\frac{d}{dt}$ would be meaningless for curves on abstract manifolds - I'm trying to understand formally what property of matrix Lie groups allows us to do this? is it the fact that they are embedded in a Euclidean space? can we do the same thing for sphere?) – newbie777 Oct 5 '19 at 19:46
• The velocity of a curve $\frac{d}{dt}\gamma$ is a tangent vector; there are several definitions which work on arbitrary manifolds. The most common one is that tangent vectors are (pointwise) derivations (i.e. pointwise directional derivatives): maps from smooth real-valued functions to real numbers satisfying $$v_p(fg)=f(p)v_p(g)+g(p)v_p(f)$$ where $f,g$ are smooth real valued functions, $p$ is a point on the manifold, and $v_p$ is a derivation at $p$. We can think of the velocity of a curve at a point as a derivation, as in the answer. – Kajelad Oct 5 '19 at 20:45
• Thanks again @Kajelad. I understand there is a more general/formal definition of tangent vectors as a derivation; i.e., in general $\frac{d}{dt} \gamma(t)$ is not even meaningful according to the definition of $\frac{d}{dt}$. But in the case of matrix Lie groups it seems that we can pretend $\gamma(t)$ maps to $\mathbb{R}^{n^2}$ and simply take the usual derivative? i.e., if $\gamma(t) \in \mathrm{SO}_n$ is a $n \times n$ matrix-valued function, I can easily compute tangent vectors to $\gamma(t)$ by explicitly computing the usual derivative of the $n \times n$ matrix-valued function wrt $t$. – newbie777 Oct 5 '19 at 21:10
• Yes. What you're effectively doing there is using the matrix elements as a set of global coordinates, with the corresponding partial derivatives as a bases for the tangent spaces. The matrix valued function $\frac{d}{dt}\gamma(t)$ is the coordinate form of the abstract tangent vector $\frac{d}{dt}\gamma$, where each element of the matrix is the component in the corresponding partial derivative. – Kajelad Oct 5 '19 at 21:48